Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{2y^2 + 4y}{y - 9} \div \dfrac{y^3 + 5y^2 + 6y}{-5y^2 + 65y - 180} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{2y^2 + 4y}{y - 9} \times \dfrac{-5y^2 + 65y - 180}{y^3 + 5y^2 + 6y} $ First factor out any common factors. $z = \dfrac{2y(y + 2)}{y - 9} \times \dfrac{-5(y^2 - 13y + 36)}{y(y^2 + 5y + 6)} $ Then factor the quadratic expressions. $z = \dfrac {2y(y + 2)} {y - 9} \times \dfrac {-5(y - 9)(y - 4)} {y(y + 2)(y + 3)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {2y(y + 2) \times -5(y - 9)(y - 4) } {(y - 9) \times y(y + 2)(y + 3) } $ $z = \dfrac {-10y(y - 9)(y - 4)(y + 2)} {y(y + 2)(y + 3)(y - 9)} $ Notice that $(y + 2)$ and $(y - 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-10y(y - 9)(y - 4)\cancel{(y + 2)}} {y\cancel{(y + 2)}(y + 3)(y - 9)} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $z = \dfrac {-10y\cancel{(y - 9)}(y - 4)\cancel{(y + 2)}} {y\cancel{(y + 2)}(y + 3)\cancel{(y - 9)}} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $z = \dfrac {-10y(y - 4)} {y(y + 3)} $ $ z = \dfrac{-10(y - 4)}{y + 3}; y \neq -2; y \neq 9 $